Given:
The quadratic equation is:
[tex]x^2-32x+54=0[/tex]
It can be written as [tex]-p=202[/tex].
To find:
The value of p in the rewritten equation.
Solution:
We have,
[tex]x^2-32x+54=0[/tex]
Isolate the constant term.
We need to make 202 on the right side. So, add 256 on both sides.
[tex]x^2-32x+256=-54+256[/tex]
[tex]x^2-32x+256=202[/tex]
[tex]-(-x^2+32x-256)=202[/tex]
Let [tex]-x^2+32x-256=p[/tex], then
[tex]-p=202[/tex]
Therefore, the value of p is [tex]-x^2+32x-256[/tex].
The given equation can be written as:
[tex]54=-x^2+32x[/tex]
Adding 148 on both sides, we get
[tex]54+148=-x^2+32x+148[/tex]
[tex]202=-(x^2-32x-148)[/tex]
Let [tex]x^2-32x-148=p[/tex], then
[tex]202=-p[/tex]
Therefore, the another possible value of p is [tex]x^2-32x-148[/tex].
